Exam (elaborations) TEST BANK FOR Physical Chemistry By Atkins 8th Edition

Exam (elaborations) TEST BANK FOR Physical Chemistry By Atkins 8th Edition (Instructor Solution Manual to ISBN 7594) Physical Chemistry, ISBN: 4334 (Instructor Solution Manual to ISBN 7594) Part 1: Equilibrium 1 The properties of gases Solutions to exercises Discussion questions E1.1(b) The partial pressure of a gas in a mixture of gases is the pressure the gas would exert if it occupied alone the same container as the mixture at the same temperature. It is a limiting law because it holds exactly only under conditions where the gases have no effect upon each other. This can only be true in the limit of zero pressure where the molecules of the gas are very far apart. Hence, Dalton’s law holds exactly only for a mixture of perfect gases; for real gases, the law is only an approximation. E1.2(b) The critical constants represent the state of a system at which the distinction between the liquid and vapour phases disappears. We usually describe this situation by saying that above the critical temperature the liquid phase cannot be produced by the application of pressure alone. The liquid and vapour phases can no longer coexist, though fluids in the so-called supercritical region have both liquid and vapour characteristics. (See Box 6.1 for a more thorough discussion of the supercritical state.) E1.3(b) The van der Waals equation is a cubic equation in the volume, V . Any cubic equation has certain properties, one of which is that there are some values of the coefficients of the variable where the number of real roots passes from three to one. In fact, any equation of state of odd degree higher than 1 can in principle account for critical behavior because for equations of odd degree in V there are necessarily some values of temperature and pressure for which the number of real roots of V passes from n(odd) to 1. That is, the multiple values of V converge from n to 1 as T → Tc. This mathematical result is consistent with passing from a two phase region (more than one volume for a given T and p) to a one phase region (only one V for a given T and p and this corresponds to the observed experimental result as the critical point is reached. Numerical exercises E1.4(b) Boyle’s law applies. pV = constant so pfVf = piVi pf = piVi Vf = (104 kPa) × (2000 cm3) (250 cm3) = 832 kPa E1.5(b) (a) The perfect gas law is pV = nRT implying that the pressure would be p = nRT V All quantities on the right are given to us except n, which can be computed from the given mass of Ar. n = 25 g 39.95 g mol−1 = 0.626 mol so p = (0.626 mol) × (8.31 × 10−2 L barK−1 mol−1) × (30 + 273K) 1.5L = 10.5 bar not 2.0 bar. 4 INSTRUCTOR’S MANUAL (b) The van der Waals equation is p = RT Vm − b − a V2m so p = (8.31 × 10−2 L barK−1 mol−1) × (30 + 273)K (1.5L/0.626 mol) − 3.20 × 10−2 L mol−1 − (1.337 L2 atm mol−2) × (1.013 bar atm−1) (1.5L/0.62¯6 mol)2 = 10.4 bar E1.6(b) (a) Boyle’s law applies. pV = constant so pfVf = piVi and pi = pfVf Vi = (1.48 × 103 Torr) × (2.14 dm3) (2.14 + 1.80) dm3 = 8.04 × 102 Torr (b) The original pressure in bar is pi = (8.04 × 102 Torr) ×
1 atm 760 Torr  ×
1.013 bar 1 atm  = 1.07 bar E1.7(b) Charles’s law applies. V ∝ T so Vi Ti = Vf Tf and Tf = VfTi Vi = (150 cm3) × (35 + 273)K 500 cm3 = 92.4K E1.8(b) The relation between pressure and temperature at constant volume can be derived from the perfect gas law pV = nRT so p ∝ T and pi Ti = pf Tf The final pressure, then, ought to be pf = piTf Ti = (125 kPa) × (11 + 273)K (23 + 273)K = 120 kPa E1.9(b) According to the perfect gas law, one can compute the amount of gas from pressure, temperature, and volume. Once this is done, the mass of the gas can be computed from the amount and the molar mass using pV = nRT so n = pV RT = (1.00 atm) × (1.013 × 105 Pa atm−1) × (4.00 × 103 m3) (8.3145 JK−1 mol−1) × (20 + 273)K = 1.66 × 105 mol and m = (1.66 × 105 mol) × (16.04 g mol−1 ) = 2.67 × 106 g = 2.67 × 103 kg E1.10(b) All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pVm/T will give the best value of R. THE PROPERTIES OF GASES 5 The molar mass is obtained from pV = nRT = m M RT which upon rearrangement gives M = m V RT p = ρ RT p The best value of M is obtained from an extrapolation of ρ/p versus p to p = 0; the intercept is M/RT . Draw up the following table p/atm (pVm/T )/(L atmK−1 mol−1) (ρ/p)/(gL−1 atm−1) 0.750 000 0..428 59 0.500 000 0..428 22 0.250 000 0..427 90 From Fig. 1.1(a),
pVm T  p=0 = 0.082 061 5 L atmK−1 mol−1 From Fig. 1.1(b),
ρ p  p=0 = 1.42755 g L−1 atm−1 8.200 8.202 8.204 8.206 0 0.25 0.50 0.75 1.0 8.20615 m Figure 1.1(a) 1.4274 1.4276 1.4278 1.4280 1.4282 1.4284 1.4286 1.4288 0 0.25 0.50 0.75 1.0 1.42755 Figure 1.1(b) 6 INSTRUCTOR’S MANUAL M = RT
ρ p  p=0 = (0.082 061 5 L atm mol−1 K−1 ) × (273.15K) × (1.42755 g L−1 atm−1 ) = 31.9987 g mol−1 The value obtained for R deviates from the accepted value by 0.005 per cent. The error results from the fact that only three data points are available and that a linear extrapolation was employed. The molar mass, however, agrees exactly with the accepted value, probably because of compensating plotting errors. E1.11(b) The mass density ρ is related to the molar volume Vm by Vm = M ρ where M is the molar mass. Putting this relation into the perfect gas law yields pVm = RT so pM ρ = RT Rearranging this result gives an expression for M; once we know the molar mass, we can divide by the molar mass of phosphorus atoms to determine the number of atoms per gas molecule M = RTρ p = (62.364 L TorrK−1 mol−1) × [(100 + 273) K] × (0.6388 g L−1) 120 Torr = 124 g mol−1 . The number of atoms per molecule is 124 g mol−1 31.0 g mol−1 = 4.00 suggesting a formula of P4 E1.12(b) Use the perfect gas equation to compute the amount; then convert to mass. pV = nRT so n = pV RT We need the partial pressure of water, which is 53 per cent of the equilibrium vapour pressure at the given temperature and standard pressure. p = (0.53) × (2.69 × 103 Pa) = 1.4¯3 × 103 Pa so n = (1.43 × 103 Pa) × (250m3) (8.3145 JK−1 mol−1) × (23 + 273)K = 1.45 × 102 mol or m = (1.45 × 102 mol) × (18.0 g mol−1 ) = 2.61 × 103 g = 2.61 kg E1.13(b) (a) The volume occupied by each gas is the same, since each completely fills the container. Thus solving for V from eqn 14 we have (assuming a perfect gas) V = nJRT pJ nNe = 0.225 g 20.18 g mol−1 = 1.115 × 10−2 mol, pNe = 66.5 Torr, T= 300K V = (1.115 × 10−2 mol) × (62.36 L TorrK−1 mol−1) × (300K) 66.5 Torr = 3.137L = 3.14 L THE PROPERTIES OF GASES 7 (b) The total pressure is determined from the total amount of gas, n = nCH4 + nAr + nNe. nCH4 = 0.320 g 16.04 g mol−1 = 1.995 × 10−2 mol nAr = 0.175 g 39.95 g mol−1 = 4.38 × 10−3 mol n = (1.995 + 0.438 + 1.115) × 10−2 mol = 3.548 × 10−2 mol p = nRT V [1] = (3.548 × 10−2 mol) × (62.36 L TorrK−1 mol−1) × (300K) 3.137L = 212 Torr E1.14(b) This is similar to Exercise 1.14(a) with the exception that the density is first calculated. M = ρ RT p [Exercise 1.11(a)] ρ = 33.5mg 250mL = 0.1340 gL−1 , p= 152 Torr, T= 298K M = (0.1340 gL−1) × (62.36 L TorrK−1 mol−1) × (298K) 152 Torr = 16.4 g mol−1 E1.15(b) This exercise is similar to Exercise 1.15(a) in that it uses the definition of absolute zero as that temperature at which the volume of a sample of gas would become zero if the substance remained a gas at low temperatures. The solution uses the experimental fact that the volume is a linear function of the Celsius temperature. Thus V = V0 + αV0θ = V0 + bθ, b = αV0 At absolute zero, V = 0, or 0 = 20.00 L + 0.0741 L◦C−1 × θ(abs. zero) θ(abs. zero) = − 20.00 L 0.0741 L◦C−1 = −270◦C which is close to the accepted value of −273◦C. E1.16(b) (a) p = nRT V n = 1.0 mol T = (i) 273.15K; (ii) 500K V = (i) 22.414 L; (ii) 150 cm3 (i) p = (1.0 mol) × (8.206 × 10−2 L atmK−1 mol−1) × (273.15K) 22.414 L = 1.0 atm (ii) p = (1.0 mol) × (8.206 × 10−2 L atmK−1 mol−1) × (500K) 0.150 L = 270 atm (2 significant figures) (b) From Table (1.6) for H2S a = 4.484 L2 atm mol−1 b = 4.34 × 10−2 L mol−1 p = nRT V − nb − an2 V 2 8 INSTRUCTOR’S MANUAL (i) p = (1.0 mol) × (8.206 × 10−2 L atmK−1 mol−1) × (273.15K) 22.414 L − (1.0 mol) × (4.34 × 10−2 L mol−1) − (4.484 L2 atm mol−1) × (1.0 mol)2 (22.414 L)2 = 0.99 atm (ii) p = (1.0 mol) × (8.206 × 10−2 L atmK−1 mol−1) × (500K) 0.150 L − (1.0 mol) × (4.34 × 10−2 L mol−1) − (4.484 L2atm mol−1) × (1.0 mol)2 (0.150 L)2 = 185.6 atm ≈ 190 atm (2 significant figures). E1.17(b) The critical constants of a van der Waals gas are Vc = 3b = 3(0.0436 L mol−1 ) = 0.131 L mol−1 pc = a 27b2 = 1.32 atm L2 mol−2 27(0.0436 L mol−1)2 = 25.7 atm and Tc = 8a 27Rb = 8(1.32 atm L2 mol−2) 27(0.08206 L atmK−1 mol−1) × (0.0436 L mol−1) = 109K E1.18(b) The compression factor is Z = pVm RT = Vm Vm,perfect (a) Because Vm = Vm,perfect + 0.12 Vm,perfect = (1.12)Vm,perfect, we have Z = 1.12 Repulsive forces dominate. (b) The molar volume is V = (1.12)Vm,perfect = (1.12) ×
RT p  V = (1.12) ×  (0.08206 L atmK−1 mol−1) × (350K) 12 atm  = 2.7 L mol−1 E1.19(b) (a) V om = RT p = (8.314 JK−1 mol−1) × (298.15K) (200 bar) × (105 Pa bar−1) = 1.24 × 10−4 m3 mol−1 = 0.124 L mol−1 (b) The van der Waals equation is a cubic equation in Vm. The most direct way of obtaining the molar volume would be to solve the cubic analytically. However, this approach is cumbersome, so we proceed as in Example 1.6. The van der Waals equation is rearranged to the cubic form V 3m−
b + RT p  V 2m +
a p  Vm − ab p = 0 or x 3 −
b + RT p  x 2 +
a p  x − ab p = 0 with x = Vm/(L mol−1 ). THE PROPERTIES OF GASES 9 The coefficients in the equation are evaluated as b + RT p = (3.183 × 10−2 L mol−1 ) + (8.206 × 10−2 L atmK−1 mol−1) × (298.15K) (200 bar) × (1.013 atm bar−1) = (3.183 × 10−2 + 0.1208) L mol−1 = 0.1526 L mol−1 a p = 1.360 L2 atm mol−2 (200 bar) × (1.013 atm bar−1) = 6.71 × 10−3 (L mol−1 ) 2 ab p = (1.360 L2 atm mol−2) × (3.183 × 10−2 L mol−1) (200 bar) × (1.013 atm bar−1) = 2.137 × 10−4 (L mol−1 ) 3 Thus, the equation to be solved is x 3 − 0.1526x 2 + (6.71 × 10−3 )x − (2.137 × 10−4 ) = 0. Calculators and computer software for the solution of polynomials are readily available. In this case we find x = 0.112 or Vm = 0.112 L mol−1 The difference is about 15 per cent. E1.20(b) (a) Vm = M ρ = 18.015 g mol−1 0.5678 g L−1 = 31.728 L mol−1 Z = pVm RT = (1.00 bar) × (31.728 L mol−1) (0.083 145 L barK−1 mol−1) × (383K) = 0.9963 (b) Using p = RT Vm − b − a V2m and substituting into the expression for Z above we get Z = Vm Vm − b − a VmRT = 31.728 L mol−1 31.728 L mol−1 − 0.030 49 L mol−1 − 5.464 L2 atm mol−2 (31.728 L mol−1) × (0.082 06 L atmK−1 mol−1) × (383K) = 0.9954 Comment. Both values of Z are very close to the perfect gas value of 1.000, indicating that water vapour is essentially perfect at 1.00 bar pressure. E1.21(b) The molar volume is obtained by solving Z = pVm RT [1.20b], for Vm, which yields Vm = ZRT p = (0.86) × (0.08206 L atmK−1 mol−1) × (300K) 20 atm = 1.059 L mol−1 (a) Then, V = nVm = (8.2 × 10−3 mol) × (1.059 L mol−1 ) = 8.7 × 10−3 L = 8.7mL 10 INSTRUCTOR’S MANUAL (b) An approximate value of B can be obtained from eqn 1.22 by truncation of the series expansion after the second term, B/Vm, in the series. Then, B = Vm
pVm RT − 1  = Vm × (Z − 1) = (1.059 L mol−1 ) × (0.86 − 1) = −0.15 L mol−1 E1.22(b) (a) Mole fractions are xN = nN ntotal = 2.5 mol (2.5 + 1.5) mol = 0.63 Similarly, xH = 0.37 (c) According to the perfect gas law ptotalV = ntotalRT so ptotal = ntotalRT V = (4.0 mol) × (0.08206 L atm mol−1 K−1) × (273.15K) 22.4L = 4.0 atm (b) The partial pressures are pN = xNptot = (0.63) × (4.0 atm) = 2.5 atm and pH = (0.37) × (4.0 atm) = 1.5 atm E1.23(b) The critical volume of a van der Waals gas is Vc = 3b so b = 13 Vc = 13 (148 cm3 mol−1) = 49.3 cm3 mol−1 = 0.0493 L mol−1 By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e., twice their radius); that volume times the Avogadro constant is the molar excluded volume b b = NA  4π(2r)3 3  so r = 1 2
3b 4πNA 1/3 r = 1 2  3(49.3 cm3 mol−1) 4π(6.022 × 1023 mol−1) 1/3 = 1.94 × 10−8 cm = 1.94 × 10−10 m The critical pressure is pc = a 27b2 so a = 27pcb 2 = 27(48.20 atm) × (0.0493 L mol−1 ) 2 = 3.16 L2 atm mol−2 THE PROPERTIES OF GASES 11 But this problem is overdetermined. We have another piece of information Tc = 8a 27Rb According to the constants we have already determined, Tc should be Tc = 8(3.16 L2 atm mol−2) 27(0.08206 L atmK−1 mol−1) × (0.0493 L mol−1) = 231K However, the reported Tc is 305.4 K, suggesting our computed a/b is about 25 per cent lower than it should be. E1.24(b) (a) The Boyle temperature is the temperature at which lim Vm→∞ dZ d(1/Vm) vanishes. According to the van der Waals equation Z = pVm RT =  RT Vm−b − a V2m  Vm RT = Vm Vm − b − a VmRT so dZ d(1/Vm) =
dZ dVm  ×
dVm d(1/Vm)  = −V 2m
dZ dVm  = −V 2m
−Vm (Vm − b)2 + 1 Vm − b + a V2m RT  = V2m b (Vm − b)2 − a RT In the limit of large molar volume, we have lim Vm→∞ dZ d(1/Vm) = b − a RT =0 so a RT = b and T = a Rb = (4.484 L2 atm mol−2) (0.08206 L atmK−1 mol−1) × (0.0434 L mol−1) = 1259K (b) By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e. twice their radius); the Avogadro constant times the volume is the molar excluded volume b b = NA  4π(2r3) 3  so r = 1 2
3b 4πNA 1/3 r = 1 2  3(0.0434 dm3 mol−1) 4π(6.022 × 1023 mol−1) 1/3 = 1.286 × 10−9 dm = 1.29 × 10−10 m = 0.129 nm E1.25(b) States that have the same reduced pressure, temperature, and volume are said to correspond. The reduced pressure and temperature for N2 at 1.0 atm and 25◦C are pr = p pc = 1.0 atm 33.54 atm = 0.030 and Tr = T Tc = (25 + 273)K 126.3K = 2.36 12 INSTRUCTOR’S MANUAL The corresponding states are (a) For H2S p = prpc = (0.030) × (88.3 atm) = 2.6 atm T = TrTc = (2.36) × (373.2K) = 881K (Critical constants of H2S obtained from Handbook of Chemistry and Physics.) (b) For CO2 p = prpc = (0.030) × (72.85 atm) = 2.2 atm T = TrTc = (2.36) × (304.2K) = 718K (c) For Ar p = prpc = (0.030) × (48.00 atm) = 1.4 atm T = TrTc = (2.36) × (150.72K) = 356K E1.26(b) The van der Waals equation is p = RT Vm − b − a V2m which can be solved for b b = Vm − RT p + a V2m = 4.00 × 10−4 m3 mol−1 − (8.3145 JK−1 mol−1) × (288K) 4.0 × 106 Pa +  0.76m6 Pa mol−2 (4.00×10−4 m3 mol−1)2  = 1.3 × 10−4 m3 mol−1 The compression factor is Z = pVm RT = (4.0 × 106 Pa) × (4.00 × 10−4 m3 mol−1) (8.3145 JK−1 mol−1) × (288K) = 0.67 Solutions to problems Solutions to numerical problems P1.2 Identifying pex in the equation p = pex + ρgh [1.4] as the pressure at the top of the straw and p as the atmospheric pressure on the liquid, the pressure difference is p − pex = ρgh = (1.0 × 103 kgm−3 ) × (9.81ms−2 ) × (0.15m) = 1.5 × 103 Pa (= 1.5 × 10−2 atm) P1.4 pV = nRT [1.12] implies that, with n constant, pfVf Tf = piVi Ti Solving for pf , the pressure at its maximum altitude, yields pf = Vi Vf × Tf Ti × pi THE PROPERTIES OF GASES 13 Substituting Vi = 4 3πr3 i and Vf = 4 3πr3 f pf =  (4/3)πr3 i (4/3)πr3 f  × Tf Ti × pi =
ri rf 3 × Tf Ti × pi =
1.0m 3.0m 3 ×
253K 293K  × (1.0 atm) = 3.2 × 10−2 atm P1.6 The value of absolute zero can be expressed in terms of α by using the requirement that the volume of a perfect gas becomes zero at the absolute zero of temperature. Hence 0 = V0[1 + αθ(abs. zero)] Then θ(abs. zero) = −1 α All gases become perfect in the limit of zero pressure, so the best value of α and, hence, θ(abs. zero) is obtained by extrapolating α to zero pressure. This is done in Fig. 1.2. Using the extrapolated value, α = 3.6637 × 10−3◦C−1, or θ(abs. zero) = − 1 3.6637 × 10−3◦C−1 = −272.95◦C which is close to the accepted value of −273.15◦C. 3.662 3.664 3.666 3.668 3.670 3.672 p /Torr Figure 1.2 P1.7 The mass of displaced gas is ρV , where V is the volume of the bulb and ρ is the density of the gas. The balance condition for the two gases is m(bulb) = ρV (bulb), m(bulb) = ρ  V (bulb) which implies that ρ = ρ . Because [Problem 1.5] ρ = pM RT the balance condition is pM = p  M  which implies that M  = p p  ×M This relation is valid in the limit of zero pressure (for a gas behaving perfectly). 14 INSTRUCTOR’S MANUAL In experiment 1, p = 423.22 Torr, p  = 327.10 Torr; hence M  = 423.22 Torr 327.10 Torr × 70.014 g mol−1 = 90.59 g mol−1 In experiment 2, p = 427.22 Torr, p  = 293.22 Torr; hence M  = 427.22 Torr 293.22 Torr × 70.014 g mol−1 = 102.0 g mol−1 In a proper series of experiments one should reduce the pressure (e.g. by adjusting the balanced weight). Experiment 2 is closer to zero pressure than experiment 1; it may be safe to conclude that M ≈ 102 g mol−1 . The molecules CH2FCF3 or CHF2CHF2 have M ≈ 102 g mol−1. P1.9 We assume that no H2 remains after the reaction has gone to completion. The balanced equation is N2 + 3H2 → 2NH3 We can draw up the following table N2 H2 NH3 Total Initial amount n n  0 n + n  Final amount n − 1 3n  0 2 3n  n + 1 3n  Specifically 0.33 mol 0 1.33 mol 1.66 mol Mole fraction 0.20 0 0.80 1.00 p = nRT V = (1.66 mol) ×  (8.206 × 10−2 L atm K−1 mol−1) × (273.15K) 22.4L  = 1.66 atm p(H2) = x(H2)p = 0 p(N2) = x(N2)p = (0.20 × (1.66 atm)) = 0.33 atm p(NH3) = x(NH3)p = (0.80) × (1.66 atm) = 1.33 atm P1.10 (a) Vm = RT p = (8.206 × 10−2 L atm K−1 mol−1) × (350K) 2.30 atm = 12.5 L mol−1 (b) From p = RT Vm − b − a V2m [1.25b], we obtain Vm = RT  p + a V2m  + b [rearrange 1.25b] Then, with a and b from Table 1.6 Vm ≈ (8.206 × 10−2 L atm K−1 mol−1) × (350K) (2.30 atm) +  6.260 L2 atm mol−2 (12.5 L mol−1)2  + (5.42 × 10−2 L mol−1 ) ≈ 28.72 L mol−1 2.34 + (5.42 × 10−2 L mol−1 ) ≈ 12.3 L mol−1 . Substitution of 12.3 L mol−1 into the denominator of the first expression again results in Vm = 12.3 L mol−1, so the cycle of approximation may be terminated. THE PROPERTIES OF GASES 15 P1.13 (a) Since B  (TB) = 0 at the Boyle temperature (section 1.3b): B  (TB) = a + b e−c/TB2 = 0 Solving for TB : TB =  −c ln −a b =  −(1131K2) ln −(−0.1993 bar−1) (0.2002 bar−1)  = 501.0K (b) Perfect Gas Equation: Vm(p, T ) = RT p Vm(50 bar, 298.15K) = 0. L bar K−1 mol−1 (298.15K) 50 bar = 0.496 L mol−1 Vm(50 bar, 373.15K) = 0. L bar K−1 mol−1 (373.15K) 50 bar = 0.621 L mol−1 Virial Equation (eqn 1.21 to first order): Vm(p, T ) = RT p (1+B  (T ) p) = Vperfect(1+B  (T ) p) B  (T ) = a + b e − c T 2 B B  (298.15K) = −0.1993 bar−1 + 0.2002 bar−1 e − 1131K2 (298.15K)2 = −0.00163 bar−1 B  (373.15K) = −0.1993 bar−1 + 0.2002 bar−1 e − 1131K2 (373.15K)2 = −0. bar−1 Vm(50 bar, 298.15K) = 0.496 L mol−1  1 − 0.00163 bar−1 50 bar  = 0.456 L mol−1 Vm(50 bar, 373.15K) = 0.621 L mol−1  1 − 0. bar−1 50 bar  = 0.599 L mol−1 The perfect gas law predicts a molar volume that is 9% too large at 298K and 4% too large at 373 K. The negative value of the second virial coefficient at both temperatures indicates the dominance of very weak intermolecular attractive forces over repulsive forces. P1.15 From Table 1.6 Tc =
2 3  ×
2a 3bR 1/2 , pc =
1 12  ×
2aR 3b3 1/2
2a 3bR 1/2 may be solved for from the expression for pc and yields
12bpc R  . Thus Tc =
2 3  ×
12pcb R  =
8 3  ×
pcVc R  =
8 3  ×  (40 atm) × (160 × 10−3 L mol−1) 8.206 × 10−2 L atm K−1 mol−1  = 210K vmol = b NA =
1 3  ×
Vc NA  = 160 × 10−6 m3 mol−1 (3) × (6.022 × 1023 mol−1) = 8.86 × 10−29 m3 vmol = 4π 3 r 3 r =
3 4π × (8.86 × 10−29 m3 ) 1/3 = 0.28 nm 16 INSTRUCTOR’S MANUAL P1.16 Vc = 2b, Tc = a 4bR [Table 1.6] Hence, with Vc and Tc from Table 1.5, b = 1 2Vc = 1 2 × (118.8 cm3 mol−1) = 59.4 cm3 mol−1 a = 4bRTc = 2RTcVc = (2) × (8.206 × 10−2 L atm K−1 mol−1 ) × (289.75K) × (118.8 × 10−3 L mol−1 ) = 5.649 L2 atm mol−2 Hence p = RT Vm − b e−a/RT Vm = nRT V − nb e−na/RT V = (1.0 mol) × (8.206 × 10−2 L atm K−1 mol−1) × (298K) (1.0L) − (1.0 mol) × (59.4 × 10−3 L mol−1) ×exp  −(1.0 mol) × (5.649 L2 atm mol−2) (8.206 × 10−2 L atm K−1 mol−1) × (298K) × (1.0L2 atm mol−1)  = 26.0 atm × e−0.231 = 21 atm Solutions to theoretical problems P1.18 This expansion has already been given in the solutions to Exercise 1.24(a) and Problem 1.17; the result is p = RT Vm  1 + b − a RT  1 Vm + b2 V2m +· · ·  Compare this expansion with p = RT Vm
1 + B Vm + C Vm2 +· · ·  [1.22] and hence find B = b − a RT and C = b 2 Since C = 1200 cm6 mol−2 , b = C 1/2 = 34.6 cm3 mol−1 a = RT (b − B) = (8.206 × 10−2 ) × (273 L atm mol−1 ) × (34.6 + 21.7) cm3 mol−1 = (22.40 L atm mol−1 ) × (56.3 × 10−3 L mol−1 ) = 1.26 L2 atm mol−2 P1.22 For a real gas we may use the virial expansion in terms of p [1.21] p = nRT V (1 + B  p +· · ·) = ρ RT M (1 + B  p +· · ·) which rearranges to p ρ = RT M + RT B  M p +· · · THE PROPERTIES OF GASES 17 Therefore, the limiting slope of a plot of p ρ against p is B  RT M . From Fig. 1.2 in the Student’s Solutions Manual, the limiting slope is B  RT M = (4.41 − 5.27) × 104 m2 s−2 (10.132 − 1.223) × 104 Pa = −9.7 × 10−2 kg−1 m3 From Fig. 1.2, RT M = 5.39 × 104 m2 s−2; hence B  = −9.7 × 10−2 kg−1 m3 5.39 × 104 m2 s−2 = −1.80 × 10−6 Pa−1 B  = (−1.80 × 10−6 Pa−1 ) × (1.0133 × 105 Pa atm−1 ) = −0.182 atm−1 B = RT B  [Problem 1.21] = (8.206 × 10−2 L atm K−1 mol−1 ) × (298K) × (−0.182 atm−1 ) = −4.4 L mol−1 P1.23 Write Vm = f (T,p); then dVm =
∂Vm ∂T  p dT +
∂Vm ∂p  T dp Restricting the variations of T and p to those which leave Vm constant, that is dVm = 0, we obtain
∂Vm ∂T  p = −
∂Vm ∂p  T ×
∂p ∂T  Vm = −
∂p ∂Vm −1 T ×
∂p ∂T  Vm = −  ∂p ∂T   Vm ∂p ∂Vm  T From the equation of state
∂p ∂Vm  T = −RT V2m − 2(a + bT )V −3 m
∂p ∂T  Vm = R Vm + b V2m Substituting
∂Vm ∂T  P = −  R Vm + b Vm2   −RT V2m − 2(a+bT ) V3m  = +  R +  b Vm   RT Vm + 2(a+bT ) V2m  From the equation of state (a + bT ) V2m = p − RT Vm Then
∂Vm ∂T  p =  R + b Vm  RT Vm + 2  p − RT Vm  =  R + b Vm  2p − RT Vm = RVm + b 2pVm − RT P1.25 Z = Vm Vom , where V om = the molar volume of a perfect gas From the given equation of state Vm = b + RT p = b + V om then Z = b + Vom Vom = 1 + b Vom For Vm = 10b, 10b = b + V om or V om = 9b then Z = 10b 9b = 10 9 = 1.11 18 INSTRUCTOR’S MANUAL P1.27 The two masses represent the same volume of gas under identical conditions, and therefore, the same number of molecules (Avogadro’s principle) and moles, n. Thus, the masses can be expressed as nMN = 2.2990 g for ‘chemical nitrogen’ and nArMAr + nNMN = n[xArMAr + (1 − xAr)MN] = 2.3102 g for ‘atmospheric nitrogen’. Dividing the latter expression by the former yields xArMAr MN + (1 − xAr) = 2.3102 2.2990 so xAr
MAr MN − 1  = 2.3102 2.2990 − 1 and xAr = 2.3102 2.2990 − 1 MAr MN − 1 = 2.3102 2.2990 − 1 39.95 g mol−1 28.013 g mol−1 − 1 = 0.011 Comment. This value for the mole fraction of argon in air is close to the modern value. P1.29 Z = pVm RT =
Tc T  ×
p pc  ×
pcVm RTc  = prV  r Tr [1.20b, 1.28] = V  r Tr  8Tr 3Vr − 1 − 3 V 2 r  [1.29] But Vr = V Vc = RTc pcVc ×
pcV RTc  = 8 3 pcV RTc [1.27] = 8 3 V  r Therefore Z = V  r Tr   8Tr 3  8V  r 3  − 1 − 3  8V  r 3 2   = V  r Tr  Tr V  r − 1/8 − 27 64(V  r )2  = V  r  1 V  r − 1/8 − 27 64Tr(V  r )2  Z = V  r V  r − 1/8 − 27 64TrV  r (2) To derive the alternative form, solve eqn 1 for V  r , substitute the result into eqn 2, and simplify into polynomial form. V  r = ZTr pr Z = ZTr/pr ZTr pr − 18 − 27 64Tr
pr ZTr  = 8ZTr 8ZTr − pr − 27pr 64ZT 2 r = 512T 3 r Z2 − 27pr × (8TrZ − pr) 64T 2 r × (8ZTr − pr)Z 64T 2 r (8ZTr − pr)Z 2 = 512T 3 r Z 2 − 216TrprZ + 27p 2 r 512T 3 r Z 3 −  64T 2 r pr + 512T 3 r  Z 2 + 216TrprZ − 27p 2 r = 0 THE PROPERTIES OF GASES 19 Z 3 −
pr 8Tr + 1  Z 2 + 27pr 64T 2 r Z − 27p2 r 512T 3 r = 0 (3) At Tr = 1.2 and pr = 3 eqn 3 predicts that Z is the root of Z 3 −
3 8(1.2) + 1  Z 3 + 27(3) 64(1.2)2 Z − 27(3)2 512(1.2)3 = 0 Z 3 − 1.3125Z 2 + 0.8789Z − 0.2747 = 0 The real root is Z = 0.611 and this prediction is independent of the specific gas. Figure 1.27 indicates that the experimental result for the listed gases is closer to 0.55. Solutions to applications P1.31 Refer to Fig. 1.3. h Ground Air (environment) Figure 1.3 The buoyant force on the cylinder is Fbuoy = Fbottom − Ftop = A(pbottom − ptop) according to the barometric formula. ptop = pbottome−Mgh/RT whereM is the molar mass of the environment (air). Since h is small, the exponential can be expanded in a Taylor series around h = 0
e−x = 1 − x + 1 2! x 2 +· · ·  . Keeping the first-order term only yields ptop = pbottom
1 − Mgh RT  20 INSTRUCTOR’S MANUAL The buoyant force becomes Fbuoy = Apbottom
1 − 1 + Mgh RT  = Ah
pbottomM RT  g =
pbottomVM RT  g = nMg  n = pbottomV RT  n is the number of moles of the environment (air) displaced by the balloon, and nM = m, the mass of the displaced environment. Thus Fbuoy = mg. The net force is the difference between the buoyant force and the weight of the balloon. Thus Fnet = mg − mballoong = (m − mballoon)g This is Archimedes’ principle. 2 The First Law: the concepts Solutions to exercises Discussion questions E2.1(b) Work is a transfer of energy that results in orderly motion of the atoms and molecules in a system; heat is a transfer of energy that results in disorderly motion. See Molecular Interpretation 2.1 for a more detailed discussion. E2.2(b) Rewrite the two expressions as follows: (1) adiabatic p ∝ 1/V γ (2) isothermal p ∝ 1/V The physical reason for the difference is that, in the isothermal expansion, energy flows into the system as heat and maintains the temperature despite the fact that energy is lost as work, whereas in the adiabatic case, where no heat flows into the system, the temperature must fall as the system does work. Therefore, the pressure must fall faster in the adiabatic process than in the isothermal case. Mathematically this corresponds toγ > 1. E2.3(b) Standard reaction enthalpies can be calculated from a knowledge of the standard enthalpies of formation of all the substances (reactants and products) participating in the reaction. This is an exact method which involves no approximations. The only disadvantage is that standard enthalpies of formation are not known for all substances. Approximate values can be obtained from mean bond enthalpies. See almost any general chemistry text, for example, Chemical Principles, by Atkins and Jones, Section 6.21, for an illustration of the method of calculation. This method is often quite inaccurate, though, because the average values of the bond enthalpies used may not be close to the actual values in the compounds of interest. Another somewhat more reliable approximate method is based on thermochemical groups which mimic more closely the bonding situations in the compounds of interest. See Example 2.6 for an illustration of this kind of calculation. Though better, this method suffers from the same kind of defects as the average bond enthalpy approach, since the group values used are also averages. Computer aided molecular modeling is now the method of choice for estimating standard reaction enthalpies, especially for large molecules with complex three-dimensional structures, but accurate numerical values are still difficult to obtain. Numerical exercises E2.4(b) Work done against a uniform gravitational field is w = mgh (a) w = (5.0 kg) × (100m) × (9.81ms−2 ) = 4.9 × 103 J (b) w = (5.0 kg) × (100m) × (3.73ms−2 ) = 1.9 × 103 J E2.5(b) Work done against a uniform gravitational field is w = mgh = (120 × 10−3 kg) × (50m) × (9.81ms−2 ) = 59 J E2.6(b) Work done by a system expanding against a constant external pressure is w = −pex V = −(121 × 103 Pa) ×
(15 cm) × (50 cm2) (100 cmm−1)3  = −91 J 22 INSTRUCTOR’S MANUAL E2.7(b) For a perfect gas at constant temperature U = 0 so q = −w For a perfect gas at constant temperature, H is also zero dH = d(U + pV ) we have already noted that U does not change at constant temperature; nor does pV if the gas obeys Boyle’s law. These apply to all three cases below. (a) Isothermal reversible expansion w = −nRT ln Vf Vi = −(2.00 mol) × (8.3145 JK−1 mol−1 ) × (22 +